I need to know the height and radius of a silo. This silo is made up of two peices. A tall cylinder, and a hemisphere that have nothing between them . the Silo must be able to hold 500000 liters of water which means the total area of the hemisphere and cylinder together are 500 cubic meters.I need to find out what dimensions would minimize the cost if construction if building the cylyndrical portion costs 300$ per square meter and the dome costs 400$ per square meter. Could someone help me figure out how i would solve this and give me the answer for the height of the cylinder and radius that would give the Silo these dimensions.
geez.. three answers with what appear to be three different answers... I am not going to solve your problem for you a lot of work for me and .. you need to learn it, BUT.. these are linear solutions.. (h r generate $) get out the graph paper and plot a simple intersect.
hemisphere uses 2*PI* R^2(m^2) of material cylinder uses 2*PI*R*H volume of cylinder = PI*R^2*H let H = kR =K*PI*R^3 volume of hemisphere = 2/3*PI*R^3 there is no way to alter the dimensions of the hemisphere as a unit. volume= (2/3* PI^R^3)+KPi*R^3 = (K+2/3)*PI*R^3 surface = 2*PI*R^2 + (2K*PI*R^2) =2(K+1)*PI*R^2 volume /surface = (K+2/3)*R/(2k+2) if we take the derivative with respect to k and set it to zero we should be able to maximize our dimensions. and then set volume as needed. R *(3K +2) -------------- 6*(k+1)
V = π r? h, => cyl. vol. V' = ? (4/3)π r?, => hemisp. vol. Total Volume: π r? h + ? (4/3) π r? = 500 m? A = 2π r h, => surface area of cyl. A' = ? 4π r? , => surface area of hemisp. Total Area: 2 π r h + 2 π r? Total Cost: $ = 300(2π r h) + 400(2π r? ) We will attempt to minimize the total cost by minimizing radius. Solve cyl. vol for h, => h = [500 - ? (4/3) π r? ] / π r? Plug into total cost equation: $ $ = 300[ 2π r {159/r? - 0.66)r ])} + 400(2 π r? ), simplify: $ = 299707/r - 1257 r? + 2513 r? = 299707/r + 1256 r? d$/dr = -299707/r? + 2(1256)r= 0, r = ?√ {299707/2512} = 4.9 m h = 159/4.9? - 0.66*4.9 = 6.56 - 3.23 = 3.33 m
What do yo know? First the total volume is 500 cubic meters. Second, you know that in order to have the lowest cost, the surface areas of the cylinder and 1/2 of the sphere must be a minimum. Third you know the formulas needed to calculate the dimensions. Volume of the cylinder = (pi)*r^2*h, Volume of half of a sphere = (1/2)*(4/3)*(pi)*r^3 Total volume = 500 Cubic Meters = Volume of cylinder + Volume of the half sphere 500 = (pi)*r^2*h + (1/2)*(4/3)*(pi)*r^3 (You need this in order to solve for h in terms of r Next you know the surface area of the cylinder portion of the silo. It is: Area Cylinder = 2*(pi)*r*h, Area of the surface area of the 1/2 sphere is: Area of sphere (1/2) = (1/2)*4*(pi)*r^2 Total Area = 2*(pi)*r*h + (1/2)*4*(pi)*r^2 = 2*(pi)*r*h + 2*(pi)*r^2 Now solve for h from the volume formula, h = (159.15/r^2) - (4/6)*r Plug this into the formula for the total area to get: Area = 2*(pi)*r*((159.15/r^2) - (4/6)*r) + 2*(pi)*r^2 = (1000/r) + 2.09*r^2 Next take the deriative of the Area with respect to r to get: Area' = (-1000/r^2) + 4.19*r, set this equal to 0 and solve for r to get: r = 6.2 meters. Now that you have r, I leave it to you to solve for h and determine the total cost. Hope this helps, Newton1Law
Let r = radius of silo, and h = the height of the cylindrical portion. V = 500 = h*pi*r^2 + (2/3)*pi*r^3, so... h = (500 - (2/3)*pi*r^3) / pi*r^2 C = 300*(h*2*pi*r) + 400*(2*pi*r^2) substitute for h, do algebra for a while C = 300*(((500 - (2/3)*pi*r^3) / pi*r^2)*2*pi*r) + 400*(2*pi*r^2) C = 300*(1000/r - (4/3)*pi*r^2) + 400*(2*pi*r^2) C = 300000/r - 400*pi*r^2 + 400*(2*pi*r^2) C = 300000/r + 400*pi*r^2 To find minimum, take dC/dr, and set to zero: dC/dr = -300000/r^2 + 800*pi*r 0 = -300000 + 800*pi*r^3 r = (3000/(8pi))^(1/3) = 4.924m h = (500 - (2/3)*pi*(4.924)^3) / pi*(4.924)^2 = 3.283m This is the same as the first respondent, so make it two votes for that being the correct answer.
